Trigonometric Identities and Equations for the ESAT

Fundamental Trigonometric Identities

Trigonometric functions can be defined in various ways. When first encountered in the context of right-angled triangles, the identity $\sin^2 \theta + \cos^2 \theta = 1$ is a direct manifestation of Pythagoras' Theorem. Consider a right-angled triangle where the hypotenuse $H = 1$, the side opposite the angle $\theta$ is $O$, and the adjacent side is $A$.

In this triangle, $\sin \theta = O/1 = O$ and $\cos \theta = A/1 = A$. Applying Pythagoras' Theorem, $A^2 + O^2 = 1^2$, which yields $\cos^2 \theta + \sin^2 \theta = 1$. While this diagram uses an acute angle, the formula is valid for any angle. This can be justified using CAST diagrams, which track the signs of trigonometric functions in all four quadrants. In these diagrams, the hypotenuse $H$ is always considered positive (often set to $1$), while $A$ and $O$ change signs depending on the quadrant.

The second fundamental identity is $\tan \theta = \frac{\sin \theta}{\cos \theta}$. This arises from the standard definitions:

$\sin \theta = \frac{O}{H}$

$\cos \theta = \frac{A}{H}$

$\tan \theta = \frac{O}{A}$

By dividing the expression for sine by the expression for cosine, we find that $\frac{\sin \theta}{\cos \theta} = \frac{O/H}{A/H} = \frac{O}{A} = \tan \theta$.

Solving Simple Trigonometric Equations

Solving equations like $\tan x = -\frac{1}{\sqrt{3}}$ for a given interval requires a systematic approach to identify all possible solutions. Whether you prefer using trigonometric graphs or CAST diagrams, the goal is to list the full set of solutions within the specified range. It is vital to ensure you do not inadvertently add or lose solutions during the process.

Equations with Transformed Arguments

When an equation involves a transformed argument, such as $2x + 60$, extra care must be taken with the interval. Consider the example:

Example: Solve $\sin^2(2x + 60) = \frac{1}{4}$ for $-360 < x < 360$

First, take the square root of both sides. It is essential to consider both the positive and negative square roots, leading to two equations:

1. $\sin(2x + 60) = \frac{1}{2}$
2. $\sin(2x + 60) = -\frac{1}{2}$

To solve the first equation, find the basic solution (the value a calculator would provide for $\sin^{-1} \frac{1}{2}$), which is $30^\circ$. This solution and the subsequent $150^\circ$ solution can be visualised on a graph or CAST diagram.

To find all values of $x$ in the range $-360 < x < 360$, we must first list all relevant solutions for the argument $(2x + 60)$ before rearranging for $x$. Because the final step involves dividing by $2$ and subtracting $60$, we must look at a wider range for the argument:

$2x + 60 = -690, -570, -330, -210, 30, 150, 390, 510, 750, 870$

Now, rearrange each value to solve for $x$:

$x = -375, -315, -195, -135, -15, 45, 165, 225, 345, 405$

Finally, filter these to keep only those within the original range $-360 < x < 360$:

$x = -315, -195, -135, -15, 45, 165, 225, 345$

Avoiding the Rearrangement Error

A common mistake is to rearrange the basic solution before finding the general solutions. If you start with $2x + 60 = 30$, rearrange to get $x = -15$, and then attempt to find other solutions for $x$, you will lose many valid results.

As seen in the graph above, this incorrect method misses the intermediate solutions generated by the higher frequency of the $2x$ function. Always find the full set of solutions for the argument first, then solve for $x$.

Using Identities to Solve Quadratic Equations

Trigonometry is often combined with quadratics. In such cases, the goal is to use identities to reach an expression involving only one trigonometric function, such as $\sin x = \text{number}$.

Example: Solve $12 \cos^2 x + 6 \sin x - 10 = 2$ for $0^\circ < x < 360^\circ$

This equation contains both $\sin$ and $\cos$. Use the identity $\cos^2 x = 1 - \sin^2 x$ to convert the equation into terms of $\sin$ only:

$12(1 - \sin^2 x) + 6 \sin x - 10 = 2$

Let $S = \sin x$ to simplify the appearance:

$12(1 - S^2) + 6S - 10 = 2$

$12 - 12S^2 + 6S - 10 = 2$

$-12S^2 + 6S = 0$

$12S^2 - 6S = 0$

Factorising the quadratic gives:

$6S(2S - 1) = 0$

This provides two possibilities: $S = \sin x = 0$ or $S = \sin x = \frac{1}{2}$. Solving these within the interval $0^\circ < x < 360^\circ$ gives the solutions $x = 30, 150, 180$. Note that $0$ and $360$ are excluded by the strict inequality of the interval.